# centralizer of sym(n)

• Jul 4th 2008, 07:01 AM
dimuk
centralizer of sym(n)
Can u help me to prove that the Z(Sym(n))={1}, centralizer of the symmetric group is equal to 1.
• Jul 4th 2008, 10:49 AM
Matt Westwood
Centraliser of symmetric group order > 3 is trivial
The proof I've got starts like this ...

The symmetric group order $n$ can be considered as the set of all permutations of $n$ letters.

Let us choose a permutation $\tau \in Z (S_n), \tau \neq e$, so that $\tau (i) = j, i \neq j$.

Since $n \ge 3$, we can find another permutation $\rho$ which interchanges j and k $(k \ne i, j)$ and leaves everything else where it is. So the inverse of $\rho$ does the same thing, and both $\rho$ and $\rho^{-1}$ leave i alone.

Thus:
$
\rho \tau \rho^{-1} (i) = \rho \tau (i) = \rho (j) = k
$

So $\rho \tau \rho^{-1} (i) = k \ne j = \tau (i)$.

If $\rho$ and $\tau$ were to commute, $\rho \tau \rho^{-1} (i) = \tau (i)$ (right multiply by $\rho$). But they don't.

Whatever $\tau \in S_n$ is, you can always find a $\rho$ such that $\rho \tau \rho^{-1} \ne \tau$.

So no non-identity elements of $S_n$ commute with all elements of $S_n$.
• Jul 4th 2008, 12:17 PM
dimuk
centralizer of Sym(n)
Thanx.
• Jul 7th 2008, 02:23 AM
dimuk
centralizer of Sym(n)
what do u mean by $\tau(i)$?
• Jul 7th 2008, 10:11 AM
Matt Westwood
In this context, $S_n$ is the set of all permutations of $n$ objects (which we can call 1, 2, 3, ..., $n$ without loss of generality).

A permutation of $n$ objects is a rearrangement of them. For example, a rearrangement of (1, 2, 3) is (1, 3, 2), and another one is (2, 1, 3), and so on.

$\tau$ is an element of $S_n$, and so it is a rearrangement of $n$ objects.

So $\tau$ applied to one of the objects of an n-element set will either move it to somewhere else or leave it where it is.

For example, if $\tau$ is the permutation that changes (1, 2, 3) to (1, 3, 2), then $\tau (1) = 1, \tau (2) = 3$ and $\tau (3) = 2$.

In this context, $\tau$ is any permutation you care to think of, but not the identity permutation (which leaves everything where it is). So there's bound to be at least one element of this n-element set which is not left where it is.

Call this element $i$.

What I mean by $\tau(i)$ is the result of applying this arbitrarily selected permutation $\tau$ to an equally arbitrarily selected element $i$, and calling it $j$.

Don't confuse $S_n$ (the set of permutations on $n$ obects) with the actual objects that $S_n$ is acting on. They are not the same thing.
• Jul 7th 2008, 11:31 AM
dimuk
centralizer of Sym(n)
thanx matt.